According to this question, $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa$, the cardinal product of all positive cardinals smaller than a given infinite cardinal $\aleph_\alpha$ equals to \begin{cases} 2^{\aleph_0}\aleph_\alpha^{|\alpha|}& \alpha \text{ is a limit ordinal}\\ 2^{\aleph_0}\aleph_\beta^{|\beta|}&\alpha=\beta+1\end{cases}
Similarly, let for every ordinal $\alpha$, $G(\alpha)=|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|$. That is, $G$ returns the cardinality of the Cartesian product of all positive ordinals smaller than the ordinal $\alpha$ it took.
Clearly, for all $0<\alpha<\omega$, $G(\alpha)=\Gamma(\alpha)=(\alpha-1)!$, and $G(\alpha)=1$ when $\alpha=0$.
For $\omega \le \alpha$, what I have gotten is that inequality:
Lemma 1 $$|\alpha|<G(\alpha) \le 2^{|\alpha|}$$
proof.
a)$$|\alpha|<|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|$$
$|\alpha|=\sum_{\xi \in Ord, \xi=0}^{\xi<\alpha} 1$
$=\sum_{\xi \in Ord, \xi=2}^{\xi<\alpha} 1$
$<\prod_{\xi \in Ord, \xi=2}^{\xi<\alpha} |\xi|$ (by Koenig's Lemma)
$=|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|$
b)$$|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi| \le 2^{\alpha}$$ $|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|$
$=\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}|\xi|$
$\le \prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}|\alpha|$
$=|\alpha|^{|\alpha|}$
$=2^{|\alpha|}$
$\Box$
Now if $\mathsf{GCH}$ holds, then $|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|=2^{|\alpha|}$ holds immediately since $|\alpha|^+=2^{|\alpha|}$. However, what about in $\mathsf{ZFC}$ but without $\mathsf{GCH}$? Can we conclude the same conclusion?